How do you express #3/ (x^2-6x+8)# in partial fractions?

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Answer 1 · 2016-07-26T12:56:26

#3/(x^2-6x+8) = 3/(2(x-4))-3/(2(x-2))#

#3/(x^2-6x+8) = 3/((x-4)(x-2)) = A/(x-4)+B/(x-2)#

Then, using Heaviside's cover up method, we find:

#A=3/(4-2) = 3/2#

#B=3/(2-4) = -3/2#

So:

#3/(x^2-6x+8) = 3/(2(x-4))-3/(2(x-2))#

#color(white)()#
Spelled out a little slower:

Given:

#3/((x-4)(x-2)) = A/(x-4)+B/(x-2)#

Multiply both sides by #(x-4)# to get:

#3/(x-2) = A+B(x-4)/(x-2)#

Then let #x=4# to find:

#3/(4-2) = A + 0#

hence #A=3/2#

Similarly find #B=-3/2#

Note that this is actually division by #0# in disguise, but is justifiable in terms of limits as #x->4# or #x->2#.